When a resistance connected in series with a cell is halved, the current is not exactly doubled but slightly less. Why???
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8
because voltage given by a cell is V = e -ir where r is internal resistance and e is electromotive force and i is current.
e = iR where R is external resistance
so
v = iR - ir
when you half R, you will get a little less than double current because i = v/(R - r) NOT v/R.
this is only because of internal resistance of battery
e = iR where R is external resistance
so
v = iR - ir
when you half R, you will get a little less than double current because i = v/(R - r) NOT v/R.
this is only because of internal resistance of battery
Answered by
1
Answer:
This is owing to the internal resistance r of the cell. The current delivered by the cell is
I=E/R+r
When external resistance R is made R/2, the current will be slightly less than 2I.
Explanation:
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